0.3x^2+0.4x=0

Solution for 0.3x^2+0.4x=0 equation:

0.3x^2+0.4x=0

a = 0.3; b = 0.4; c = 0;
Δ = b2-4ac
Δ = 0.42-4·0.3·0
Δ = 0.16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.4)-\sqrt{0.16}}{2*0.3}=\frac{-0.4-\sqrt{0.16}}{0.6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.4)+\sqrt{0.16}}{2*0.3}=\frac{-0.4+\sqrt{0.16}}{0.6}
$